Koliko se atoma azota nalazi u 280 ml azot(I)oksida(normalni uslovi)?
n(N2O)=280l/1000/22.4dm^3/l=0.0125mol n(N)=2*n(N2O)=0.025mol N=n(N)*Na=0.025mol*6*10^23mol^-1=1.5*10^22 atoma azota.
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n(N2O)=280l/1000/22.4dm^3/l=0
n(N2O)=280l/1000/22.4dm^3/l=0.0125mol
n(N)=2*n(N2O)=0.025mol
N=n(N)*Na=0.025mol*6*10^23mol^-1=1.5*10^22 atoma azota.